3.6 Hamiltonian systems

In classical mechanics, free vibrations of a conservative system can be described by

 

M q'' + K(q) = 0 (3.24)

where the mass M is a symmetric, positive definite matrix and the restoring forces represented by the vector K(q) have nonlinear elements expressed as smooth functions of the generalized co-ordinates q_i, $i = 1,2, \ldots , n$. Assume, without loss of generality, that M=I and $K(q) = \nabla V(q)$, where V(q) is the potential energy of the system. Then the equation of the motion (3.24) can be expressed as

$$q'' + \nabla V (q) = 0 \, .$$ (3.25)

The kinetic energy of the system is given by

$$T = \frac{1}{2} \langle q', q' \rangle$$

and the total energy H=T+V, i.e.

$$H (q,q') = \frac{1}{2} \langle q', q' \rangle + V(q)$$ (3.26)

is called the Hamiltonian.

If we set p = q', then it is easy to see, from (3.25) and (3.26), that

$$\left\{ \begin{array}{l} q' = {\dss\frac{\partial H}{\partial... ...p' = - {\dss\frac{\partial H}{\partial q}}, \end{array}\right.$$ (3.27)

which is called Hamilton's equations for the mechanical system.

Let

$$x = {q \choose p} \, , \quad L = \left[ \begin{array}{rl} 0 & 1 \\ -1 & 0 \end{array}\right] \, .$$

Then (3.27) can be written as

$$x' = L\nabla H(x) \, .$$ (3.28)

The equilibrium points of (3.28) are given by

$$\left\{ \begin{array}{l} {\dss\frac{\partial H}{\partial q}} ... ... {\dss\frac{\partial H}{\partial p}} = 0, \end{array} \right.$$

which implies

$$p = 0 \quad \mbox{and}\quad \nabla V(q) = 0 \, ,$$

i.e. $\ol{x} = (\ol{q}, 0)$, where $\nabla V(\ol{q}) = 0$.

Thus is we take H as a Lyapunov function, then it is easy to see, by Theorem 3.5.1, that the equilibrium $\ol{x} = (\ol{q}, 0)$ is stable if the potential energy function V(q) is positive definite with respect to $\ol{q}$. This establishes the famous Lagrange's theorem. An equilibrium state $\ol{x} = (\ol{q}, 0)$ is stable if the potential energy has local minimum at $\ol{q}$.

This is a result which is rather expected from physical intuition.

Example 3.6.1:     Marble moving along landscapes

Marble rolling back and forth about $\ol{q}$.

In general, the potential minima yield stable equilibria and maxima yield unstable (hyperbolic) equilibria. >

Example 3.6.2:     Consider the mass-spring system subject to a force $F(q) = - kq + \alpha q^{3}$, k > 0, $\alpha \in R$.

Here $\alpha < 0$ corresponds to a ``hard'' spring, i.e., $F \sim - q^{3}$ as $q \raro \infty$, whereas $\alpha > 0$ corresponds to a ``soft'' spring, i.e. the restoring force becomes zero at $x = \pm \sqrt{k/\alpha}$. The potential energy is, for general $\alpha$,

$$V (q) = \frac{1}{2} k q^{2} - \frac{1}{4} \alpha q^{4}$$

and plotted below for both cases.

(0,0) is stable, $\left( \pm \sqrt{k/\alpha}, 0 \right)$ are unstable.

$H(q,p) = \frac{1}{2} p^{2} + \frac{1}{2} k q^{2} - \frac{1}{4} \alpha q^{4}$

          
Example 3.6.3:     Consider the undamped pendulum

$$\left\{ \begin{array}{l} x'_{1} = x_{2}, \\ x'_{2} = - {\dss\frac{g}{\ell}} \sin x_{1} \, .\end{array}\right.$$

The potential energy $V(x_{1}) = \frac{g}{\ell} (1-\cos x_{1})$ is periodic with minima at $x_{1} = 2n \pi$ and maxima at $x_{1} = (2n + 1) \pi$, $n = 0, \pm 1, \pm 2, \ldots$ .

Thus $(2n \pi, 0)$ are stable and $((2n+1) \pi , 0 )$ are unstable. A complete understanding of the phase portraits for this problem is made possible with the Hamiltonian function

$$H(x_{1}, x_{2}) = \frac{1}{2} x_{2}^{2} + \frac{g}{\ell} (1- \cos x_{1}) \, .$$

Note that periodic orbits have energies $0 \leq H < 2 \frac{g}{\ell}$, orbits which travel from one unstable equilibrium to another have energies $H = 2 \frac{g}{\ell}$ and trajectories with energy $H > 2 \frac{g}{\ell}$ move above or below the x₁-space. The trajectories connecting the saddle points at $((2n+1) \pi , 0 )$ are called separatrices since they separate regions of phase space with qualitatively different dynamical behaviour.

Example 3.6.3 Simulation

From the preceding discussion, you may guess that if the DE describes a physical system, then the choice of Hamiltonian or total energy as a Lyapunov function will always work. Unfortunately, this is somewhat incorrect, especially when we question what ``always work'' means. The following example should illustrate the idea.

Example 3.6.4:     Consider the damped linear oscillator

$$\left\{ \begin{array}{ll} x' = y, & \\ y' = - kx - \alpha y , & \alpha = 1, \quad k = \frac{1}{2} \, . \end{array} \right.$$

The total energy function is $H(x,y) = \frac{1}{2} y^{2} + \frac{1}{4} x^{2}$ and

$$\dot{H} (x,y) = y \left( - \frac{1}{2} x - y \right) + \frac{1}{2} xy = - y^{2} \leq 0 \, .$$

Note that the extra ``equals'' sign implies that H is not a strict Lyapunov function, hence we can deduce only that (0,0) is stable. But $A = \left[ \begin{array}{rr} 0 & 1 \\ - \frac{1}{2} & -1 \end{array} \right]$, $\lambda = - \frac{1}{2} \pm \frac{i}{2}$ which implies that (0,0) is asymptotically stable.

The question arises: ``Can't we do better with the Lyapunov approach?''

The answer is ``yes'', only that we should choose a ``more suitable'' Lyapunov function. To ascertain what is ``more suitable'', we look at where the problem arose above.

The problem $\dot{H} = 0$ occurs only for y=0. If we look at the picture below, we see that, on the line y=0, the direction field, $(x', y') = \left( 0, - \frac{1}{2} x \right)$, points only in the y direction, but $\nabla H = \left( \frac{1}{2} x, 0 \right)$ points along the x-axis for y =0. The problem then, lies in the fact that the ellipses defined by level curves of
$H(x,y) = \frac{1}{2} y^{2} + \frac{1}{4} x^{2}$ have their principal axes lying along the x and y axis.

Note that the orthogonality of $\nabla H$ and f occurs only on the x-axis and the solution curves point inward into the ellipses at other places. Now it appears that we could circumvent the problems on the line y=0 by ``tilting'' the ellipses, i.e. by considering

$$V (x,y) = ax^{2} + bxy + cy^{2} \, .$$

It turns out $a = \frac{1}{2}$, $c = \frac{1}{2}$ and $b = \frac{1}{2}$. Thus

$$\dot{V} = xy + \frac{1}{2} y^{2} + \frac{1}{2} x \left( - \fr... ...} x - y \right) = - \frac{1}{4} x^{2} - \frac{1}{2} y^{2} \, .$$

V is a strict Lyapunov function and (0,0) is asymptotically stable.

Example 3.6.4 Simulation

Remark:     We conclude from the above example that the mechanical energy function is not the best Lyapunov function. Indeed, it was the energy function for mechanical systems which inspired Lyapunov's stability method. But the latter have exhibited much greater power than the former method and it can be applied to very general systems. Perhaps, the applicability of this method today has gone beyond the imagination of Lyapunov himself.